# Product of reflection matrices without eigenvalue 1

1. Jul 17, 2010

### zpconn

I'm wondering if anybody could suggest some techniques that might be brought to bear on the following problem:

Suppose a finite sequence $$M_1,M_2,\dots,M_k$$ of $$4\times 4$$ orthogonal reflection matrices is given. I'm interested in determining conditions on these matrices that will guarantee that the product of any ordered sub-collection having an even and nonzero number of elements does not have eigenvalue 1.

This is really the same as asking whether the result will be a simple or "double" rotation. If it's a simple rotation, the product will have an axis-plane that is fixed point-by-point and so will have eigenvalue 1. If it's a "double" rotation (two independent rotations in orthogonal 2-planes), then it will not have eigenvalue 1.

Or better yet: does anybody have any ideas for generating a list of matrices satisfying the specified property (besides just randomly generating matrices and testing all generated collections)?

2. Jul 17, 2010

### zpconn

Actually, I now don't think any such lists of 4x4 matrices exist.

Claim: Given at least three reflections of $$R^4$$ in distinct hyperplanes through the origin, the composition of some two of them is a simple rotation fixing a two-dimensional plane through the origin(its axis).

Proof(?): Maybe there's a proof using matrices directly, but my method uses the following connection with quaternions: the reflection of $$R^4$$ in the homogenous hyperplane orthogonal to the unit quaternion $$u$$ is the map $$q \mapsto -qvq$$ where $$v$$ is the conjugate of $$u$$. (I can't get the \overline command in LaTeX to work here for some reason, so I will write the conjugate of $$u_i$$ as $$v_i$$ from here on out.) Another nontrivial observation, which follows from the previous one, is that the orthogonal rotations of $$R^4$$ are precisely the quaternion maps $$q \mapsto aqb$$ for unit quaternions $$a,b$$.

Suppose three unit quaternions $$u_1,u_2,u_3$$ are given. The previous two observations imply [snip] that $$u_1 = u_2 = (v_1)^{-1}$$, and that's the end of the proof.

Now, my vague understanding is that in higher dimensions the quaternions are replaced by the so-called Clifford algebras. Most sources I've looked at for Clifford algebras are a little too advanced for me. Does anybody know of a good, easy introduction to Clifford algebras?